Diastereocontrol in Acyclic systems 2

Question 1

What is the lowest energy conformation in propene

Answer 1

1. A C-H bond eclipses the C=C group (inside position- where torsion angle =0 degrees to C=C)

Question 2

Why is propene most stable when C-H bond is in the inside position

Answer 2

1. Gives the best orbital alignment for two stabilising sigmaC-H –>piC-C hyperconjugative interactions (seen in acetaldehyde)
2. C=C bond is electron-rich, so also get piC-C –>sigma*C-H negative hyperconjugative interaction (similar to anomeric effect)
3. Electrons end up lower in energy overall- fill lower energy MO.

Question 3

What would happen if the allylic H was at the outside position to the C=C

Answer 3

1. Allylic H lies in the outside position 180 Degrees from C=C
2. Destabilising sigma C-H –> pi C-C interactions occur between the C-H bonds and the pi-orbital of the C=C group putting energy at a maximum.

Question 4

What is hyperconjugation

Answer 4

1. Hyperconjugation is the stabilising interaction that results from the interaction of the electrons in a σ-bond (usually C-H or C-C) with an adjacent empty or partially filled p-orbital or a π-orbital
2. Gives an extended molecular orbital that increases the stability of the system.

Question 5

What is most stable conformation of but-1-ene

Answer 5

1. A conformation with hydrogen in the inside position is lower in energy than that with methyl
2. Then conformation where ME is in inside position
3. Then conformation where H or Me are in outside position

Question 6

What is but-1-ene conformation with hydrogen in inside position more stable

Answer 6

1. Same explanation as propene
2. Additional steric interaction known as 1,3-allylic strain (A^1,3 strain)
3. An energy raising effect
4. If methyl was in inside position it would interact with the H on opposite side of C=C- sterics not good
5. Small effect

Question 7

What is most stable conformation of (Z)-pent-2-ene

Answer 7

1. Where the Me is perpendicular- 90 degrees to the C=C
2. Then where H is in inside position
   3 Then where Me is in inside position - very high compared to the other 2

Question 8

Why is the most stable conformation of (Z)-pent-2-ene where Me is in the outside

Answer 8

1. Increasing the size of the eclipsing substituent on the alkene massively increases the effect of the 1,3-allylic strain
2. Now 2 Me groups interacting when Me in inside position as Me on otherside of C=C as well - big strain
3. No strain when Me in 90 degrees so most stable

Question 9

What is most stable conformation of 2-methyl-2-butene

Answer 9

1. When H is in the inside position and Me facing up
2. Then Me in inside position- slight A^ 1,3
3. Then when Me is in outside position
4. Highest energy is H in outside position

Question 10

Why is the most stable conformation of 2-methyl-2-butene when H is in inside position

Answer 10

1. Same reasons as previously
2. Me in inside- slight sterics
3. H in outside- 1,2-allylic strain A^1,2 strain
4. Me in outside- Destabilising sigma C-H –> pi C-C - 1,2-allylic strain A^1,2 strain
5. This strain severely destabilises conformation - 2 Me groups interacting

Question 11

Is the reactive conformation the same as the ground state conformation

Answer 11

1. Not always

Question 12

What is a general requirement for a substrate conformation for a reaction- Alkene and electrophile

Answer 12

1. A sigma bond at the allylic position is aligned antti to the forming bond in order to stabilise the TS
2. Not single transition state as with C=O additions with nucleophiles as approach trajectories depend on both electrophile and alkene

Question 13

What is the main concern for electrophilic additions to alkenes where none of the allylic substituents are electronegative atoms

Answer 13

1. Steric interactions as like simple F-A model for C=O additions
2. TS with the least allylic strain and minimal steric interactions between the alkene and electrophile will be the lowest in energy

Question 14

What reactions can 1,3-Allylic strain act as a stereocontrol feature

Answer 14

1. Epoxidation
2. Hydroboration
3. Iodolactonisation
4. Substrate-directed reactions

Question 15

Describe how 1,3-Allylic strain act as a stereocontrol feature in epoxidation

Answer 15

1. Small group is in inside position of alkene to minimise the A^1,3 strain and larger group pointing upish
2. m-CPBA approaches anti to large group
3. Epoxide group forms same side to small group and opposite to large
4. In general Z-substituted alkene is required for high diastereoselectivity as A^1,3 strain is much more severe
5. dr increases when R group on opposite side of C=C is larger

Question 16

Describe how 1,3-Allylic strain act as a stereocontrol feature in hydroboration

Answer 16

1. Large group in anti position (90 degrees to C=C) - avoid clash with electrophile
2. Small group in inside position to minimise A^1,3 strain
3. BH3 approaches anti to large group
4. B adds to the less substitutes end of the alkene (anti-Markovnikov)

Question 17

Describe how 1,3-Allylic strain act as a stereocontrol feature in iodolactonisation

Answer 17

1. Small group in inside position to minimise A^1,3 strain
2. Get rapid and reversible iodonium ion formation - only productive when reacts on top face
3. Iodolactonization via SN2 - forms 6 ring intermediate
4. All groups in chair like TS should be equatorial

Question 18

Describe how 1,3-Allylic strain act as a stereocontrol feature in hydroxyl-directed epoxidations

Answer 18

1. Same process as hydroxyl-directed epoxidation of cyclic allylic alcohols
2. Z-substituted alkene generally required for high diastereoselectivity
3. OH in same side as O e.g both pointing forward

Question 19

Describe how 1,3-Allylic strain act as a stereocontrol feature in substrate directed hydrogenations

Answer 19

1. Using cationic iridium or rhodium catalysts
2. Addition on top face
3. Small group in inside position OH pointing up to allow interaction with Rh

Question 20

What reactions can be controlled using 1,2-allylic strain

Answer 20

1. When 2 non-hydrogen substituents attached to atoms 1 and 2
2. Hydroboration

Question 21

How is hydroboration controlled by 1,2-allylic strain

Answer 21

1. Hydroboration of chiral allylic alcohols is reagent-dependent
2. For BH3- original OH will be on same side as R group
3. 9-BBN-H opposite sides

Question 22

Describe why BH3 addition to a chiral allylic alcohol is syn-selective

Answer 22

1. Medium sized OH is in inside position minimises A^1,2 strain to methyl on opposite side
2. L group is anti- borane approaches anti to this group
3. B adds to less-substituted end of the alkene

Question 23

Describe why 9-BBN-H is anti-selective to chiral allylic alchols

Answer 23

1. Borane is so large that it encounters severe non-bonded steric interactions with the inside substituent so it is preferable to put the smallest group (H) in this position despite the increase in A^1,2 strain

Question 24

What preferences to electrophilic additions to C=C display when it comes to donor/acceptor groups at the allylic position

Answer 24

1. Essentially opposite to nucleophilic additions to C=O
2. Donor groups (electropositive) prefer anti position
3. Acceptor groups prefer inside position (electronegative) - but exceptions

Question 25

What are exceptions to the polar generalised rules

Answer 25

1. Some substrate-directed or intramolecular reactions
2. Were conformational effects may override the above stereoelectronic preferences

Question 26

What is the inside alkoxy effect

Answer 26

1. The preference of allylic electronegative groups to occupy the inside position of the allyl sustem so as to maximise the reactivity (HOMO energy) of the alkene

Question 27

What is conformational stereocontrol

Answer 27

1. If a reaction involves formation of one or more 6-membered rings, then conformational stereocontrol can be operative
2. For 6-membered ring the most stable TS is chair-like with as many substituents psuedoequatorial as possible
3. Shown in cationic polycyclisations in the lab